lim(sqx^2+2 - sqx^2-6x) for x->infinity

Expert Answers

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I think you mean `sqrt(x^2+2)-sqrt(x^2-6x).`

To remove the radicals, let us multiply the top and the bottom of this expression by the sum of the square roots. This will remove the square roots from the numerator.


This is equal to


Now if we multiply the top and the bottom by 1/x we can simplify this into something we can take the limit of.

`(6x+2)/(sqrt(x^2+2)+sqrt(x^2-6x))=(1/x(6x+2))/(1/x (sqrt(x^2+2)+sqrt(x^2-6x)))=(6+2/x)/(sqrt((x^2+2)/x^2)+sqrt((x^2-6x)/x^2))`

It is in turn equal to  `(6+2/x)/(sqrt(1+2/x^2)+sqrt(1-6/x))`

and here all terms have limits:

`(6+0)/(sqrt(1+0)+sqrt(1-0))=3` when `x->+oo,` and `-(6+0)/(sqrt(1+0)+sqrt(1-0))=-3` when `x->-oo.`



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